----------------
/ 18000 x 168 x 4 ------
d = / --------------- = / 268.8 = 16, inches nearly.
\/ 5000 x 9 \/


[TeX: $d = \sqrt{\frac{1800 \times 168 \times 4}{5000 \times 9}}
= \sqrt{268.8} = 16$]

Case III. Given the safety load 22,400 lbs.
the depth 18 inches.
the length 14 feet.

Required the breadth.
Deriving b from the foregoing, we have,

W x 4L
b = ----------
5000 x d²

[TeX: $b = \frac{W \times 4L}{5000 \times d^2}$]

substituting

22400 x 4 x 168
b = --------------- = 9.3 inches nearly.
5000 x 324

[TeX: $b = \frac{22400 \times 4 \times 168}{5000 \times 324} = 9.3$]